3 Answers
3

The sum of the first $n$ natural numbers is well-known to be
$$
\sum_{k=1}^nk=\frac{n(n+1)}{2}.
$$
Thus the expression within the brackets can be written as
$$
\left[\frac{n(n+1)}{2}+2\frac{(n-1)n}{2}+3\frac{(n-2)(n-1)}{2}+\cdots +n\right]
$$
or in sum-notation as
$$
\sum_{k=1}^n\left(k\frac{(n+1-k)(n+2-k)}{2}\right).
$$
Now multiply out the parenthesis in the numerator and sum each term seperately - this only requires knowledge of the sums:
$$
\sum_{k=1}^n k^2=\frac{n(n+1)(2n+1)}{6},\qquad \sum_{k=1}^n k^3=\frac{n^2(n+1)^2}{4}.
$$
Finally, divide each term by $n^4$ and let $n\to\infty$.

Now consider that only those terms with $n^4$ will have a non-zero limit (all sums with powers $l^p$ will contribute a term with $n^{p+1}$). All lower order terms have a limit of zero. This means the limit can be written as